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In the reaction: Na(2)S(2)O(3)+4Cl(2)+5H...

In the reaction: `Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl`, the equivalent weight of `Na_(2)S_(2)O_(3)` will be: (M= molecular weight of `Na_(2)S_(2)O_(3))`

A

M/4

B

M/8

C

M/1

D

M/2

Text Solution

Verified by Experts

The correct Answer is:
B
`Na_(2)overset(+2)S_(2)O_(3)rarrNa_(2)overset(+6)SO_(4)`
the total change in oxidation number `=4xx2=8`
`:. E_(Na_(2)S_(2)O_(3))=(mol.wt.)/(V.f)=(M)/(8)`
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In the section :Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl . the equivalent weight of Na_(2)S_(2)O_(3) will be : (M= molecular weight of Na_(2)S_(2)O_(3) )

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Knowledge Check

  • In the section :Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl . the equivalent weight of Na_(2)S_(2)O_(3) will be : (M= molecular weight of Na_(2)S_(2)O_(3) )

    A
    `M//4`
    B
    `M//8`
    C
    `M//1`
    D
    `M//2`

  • In the reaction 2Na_(a)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+ 2NaI , the equivalent weight of Na_(2)S_(2)O_(3) (mol. Wt. =M) is equal to

    A
    (a)`M`
    B
    (b)`M//2`
    C
    (c )`M//3`
    D
    (d)`M//4`

  • Na_(2)SO_(3)+H_(2)O_(2) rarr Na_(2)SO_(4) +H_(2)O , in reaction

    A
    `H_(2)O_(2)` is bleached
    B
    `H_(2)O_(2)` is reduced
    C
    `H_(2)O_(2)` is dehydrated
    D
    `H_(2)O_(2)` is neither oxidised nor reduced
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